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In tossing four fair dice, what is the probability of tossing, at most one 3?

This is the beginning of the solution I have:

The number of outcomes of tossing four dice is $6 \times 6 \times 6 \times 6 =6^4$.

Outcomes of getting at most one 3 = outcomes of getting no 3 $ +$ outcomes of getting one 3.

The number of outcomes of getting no 3 is $5 \times 5 \times 5 \times 5 =5^4$.

The number of outcomes of getting one 3 is $4 \times 5^3$.

But I don't understand why it's $4\times 5^3$. Can any one assist?

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    $\begingroup$ There are $4$ places you might have gotten the $3$ and $5$ choices for each of the other $3$ places. $\endgroup$ – lulu Feb 28 '19 at 16:25
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To get the number of outcomes with exactly one 3:

  1. Choose one tossing which is going to result in a 3. This can be done in $\binom{4}{1}$ ways because there are 4 tossings in total.
  2. Since the other tossings must not contain a 3, there are 5 numbers available for them. Since there are 3 tossings left, the number of ways is $5^3$

Using the multiplicative property you get $4 \cdot 5^3$.

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    $\begingroup$ Like the permutation 4 C 1? $\endgroup$ – ErinA Feb 28 '19 at 16:31
  • $\begingroup$ @ErinA Yes, indeed. $\endgroup$ – Haris Gušić Feb 28 '19 at 16:33
  • $\begingroup$ @ErinA Have I answered your question? $\endgroup$ – Haris Gušić Jun 24 '19 at 14:42

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